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## Q. 3.5

Two plane parallel plates A and B are placed 3 mm apart and potential of B is made 200 V positive with respect to plate A. An electron starts from rest from plate A. Calculate (i) the velocity of the electron on reaching plate B, (ii) time taken by the electron to travel from plate A to plate B, and (iii) kinetic energy of the electron on reaching the plate B.

## Verified Solution

(i) The electron starts from rest at plate A, therefore, the initial velocity is zero. The velocity of the electron on reaching plate B is,

$v=\sqrt{\frac{2 qV}{m} }=\sqrt{\frac{2(1.602\times 10^{-19} )200}{(9.1\times 10^{-31} )} }=8.38\times 10^{6}m/s$

(ii) Time taken by the electron to travel from plate A to plate B can be calculated from the average velocity of the electron in transit. The average velocity is

$v_{average}=\frac{Initial velocity+ Final velocity}{2}=\frac{0+8.38\times 10^{6} }{2}=4.19\times 10^{6}m/s$

Therefore, the time taken for travel is

$Time =\frac{Separation between the plates}{Average velocity}=\frac{3 \times 10^{-3} }{4.19\times 10^{6} }=0.71\times 10^{-9}s.$

(iii) Kinetic energy of the electron on reaching the plate B is

Kinetic energy = q V = $(1.602 × 10^{–19}) 200 = 3.2 × 10^{–17} J$